Tip:
Highlight text to annotate it
X
- WE WANT TO FIND THE INTEGRAL OF THE QUANTITY 4X + 5
DIVIDED BY THE QUANTITY X SQUARED + 4.
IN IT'S CURRENT FORM THOUGH NONE OF OUR INTEGRATION FORMULAS
SEEM TO APPLY, BUT BECAUSE WE HAVE A SUM IN THE NUMERATOR HERE
WE CAN WRITE THIS AS A SUM OF TWO DIFFERENT INTEGRALS.
THIS IS EQUAL TO THE INTEGRAL OF 4X
DIVIDED BY THE QUANTITY X SQUARED + 4 + THE INTEGRAL
OF 5 DIVIDED BY THE QUANTITY X SQUARED + 4.
NOW FOR THE NEXT STEP, LET'S GO AHEAD AND FACTOR OUT
THE FOUR AND FACTOR OUT THE FIVE.
SO WE'D HAVE 4 x INTEGRAL OF X DIVIDED BY X SQUARED + 4 + 5
x THE INTEGRAL OF 1 DIVIDED BY X SQUARED + 4.
AND LOOKING AT THIS FIRST INTEGRAL,
THIS DOES FIT THE FORM OF THE INTEGRAL OF 1/U DU.
NOTICE IF WE LET U EQUAL THE DENOMINATOR OF X SQUARED + 4
THEN DIFFERENTIAL U, OR DU, IS EQUAL TO 2X DX.
NOTICE HOW WE HAVE X DX.
SO IF WE DIVIDE BOTH SIDES BY 2 WE CAN SAY 1/2DU = X DX.
NOTICE THE SECOND INTERVAL FITS THIS FORM HERE
WHERE THE DENOMINATOR WOULD BE "A" SQUARED + U SQUARED
WHERE "A" SQUARED IS OUR CONSTANT.
SO LET'S REWRITE THIS AS 4 + X SQUARED
INSTEAD OF X SQUARED + 4.
IN THIS FORM WE SHOULD RECOGNIZE "A" SQUARED EQUALS FOUR
SO "A" EQUALS 2 AND U SQUARED EQUALS X SQUARED
SO U IS JUST EQUAL TO X.
SO IF U IS EQUAL TO X THEN DU EQUALS DX,
THEREFORE NO U SUBSTITUTION IS REQUIRED.
LET'S REWRITE THIS FIRST INTEGRAL IN TERMS OF U.
WE'D HAVE 4 x THE INTEGRAL OF OUR DENOMINATOR
WOULD JUST BE U AND THEN X DX IS EQUAL TO 1/2DU.
WE'LL FACTOR OUT THE 1/2 AND WE HAVE 1DU HERE.
AND NOW FOR THE SECOND INTEGRAL WE'LL JUST WRITE THIS AS
+ 5 x THE INTEGRAL OF 1 DIVIDED BY--IF WE WANT WE CAN WRITE 4
AS 2 SQUARED AND + X SQUARED.
NOW WE CAN INTEGRATE.
THIS WILL BE 2 x INTEGRAL OF 1/U DU,
THAT'D BE NATURAL LOG ABSOLUTE VALUE + C OR NATURAL LOG--
WE DON'T NEED THE ABSOLUTE VALUE THOUGH BECAUSE X SQUARED + 4
IS U, WHICH WOULD ALWAYS BE POSITIVE.
SO WE JUST HAVE X SQUARED + 4 + C SUB 1
OUR CONSTANT OF INTEGRATION.
AND THEN WE HAVE + 5 x 1/A,
WHICH WOULD BE 1/2 OR 1/2 ARCTANGENT OF U DIVIDED BY "A",
WHICH IS X DIVIDED BY 2 + C.
WE'LL CALL IT + C SUB 2.
NOW LET'S GO AHEAD AND DISTRIBUTE.
WE'LL HAVE 2 NATURAL LOG OF X SQUARED + 4 + 2 x C SUB 1,
WHICH WE'LL COME BACK TO,
+ 5/2 ARCTANGENT OF X DIVIDED BY 2 + 5 x C SUB 2.
WELL, 2 x C SUB 1 AND 5 x C SUB 2, IT'S JUST ANOTHER CONSTANT.
SO IF WE LET C EQUAL 2 x C SUB 1 + 5 x C SUB 2,
WE CAN JUST PUT + C HERE.
THIS WOULD BE OUR ANTI-DERIVATIVE.
I HOPE YOU FOUND THIS HELPFUL.